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arsen [322]
3 years ago
13

Which type of bond has an electronegativity difference greater than 1.7? a. ionic c. nonpolar covalent b. metallic d. polar cova

lent
Physics
1 answer:
cestrela7 [59]3 years ago
8 0

Answer: Option (a) is the correct answer.

Explanation:

An ionic bond is formed by transfer of electrons between the two chemically combining atoms. Whereas a covalent bond is defined as the bond formed by sharing of electrons between the two chemically combining atoms.

When electronegativity difference is from 0.0 to 0.4 then bond formed between the two atoms is non-polar covalent in nature.

When electronegativity difference is greater than 0.4 and less than 1.7 then bond between the two atoms is a polar covalent bond.

When electronegativity difference is 1.7 or greater than the bond formed is ionic in nature.

Thus, we can conclude that ionic type of bond has an electronegativity difference greater than 1.7.

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25 POINTS ------------------ (usatestprep)
Airida [17]

Answer:

D) The heavier ball will have a higher temperature because the change of temperature is inversely proportional to mass.

Explanation:

As stated in the problem, the amount of heat released by each ball is

Q=mC_p \Delta T

where

m is the mass of the ball

Cp is the specific heat of iron (so, it is equal for both balls)

\Delta T is the change in temperature of each ball

In this problem, we are said that the amount of heat released by the two balls, Q, is the same. Cp is also the same: this means that the product m\Delta T must be the same for the two balls. So, the mass and the change in temperature are inversely proportional: therefore, the heavier ball will have a smaller change in temperature. And since both balls starts from the same temperature, 100 C, this means that the heavier ball will reach a higher temperature than the lighter ball.

6 0
3 years ago
An open container holds ice of mass 0.555 kg at a temperature of -16.6 ∘C . The mass of the container can be ignored. Heat is su
s2008m [1.1K]

Answer: A. 23.59 minutes.

              B. 249.65 minutes

Explanation: This question involves the concept of Latent Heat and specific heat capacities of water in solid phase.

<em>Latent heat </em><em>of fusion </em>is the total amount of heat rejected from the unit mass of water at 0 degree Celsius to convert completely into ice of 0 degree Celsius (and the heat required for vice-versa process).

<em>Specific heat capacity</em> of a substance is the amount of heat required by the unit mass of a substance to raise its temperature by 1 kelvin.

Here, <u>given that</u>:

  • mass of ice, m= 0.555 kg
  • temperature of ice, T= -16.6°C
  • rate of heat transfer, q=820 J.min^{-1}
  • specific heat of ice, c_{i}= 2100 J.kg^{-1}.K^{-1}
  • latent heat of fusion of ice, L_{i}=334\times10^{3}J.kg^{-1}

<u>Asked:</u>

1. Time require for the ice to start melting.

2. Time required to raise the temperature above freezing point.

Sol.: 1.

<u>We have the formula:</u>

Q=mc\Delta T

Using above equation we find the total heat required to bring the ice from -16.6°C to 0°C.

Q= 0.555\times2100\times16.6

Q= 19347.3 J

Now, we require 19347.3 joules of heat to bring the ice to 0°C  and then on further addition of heat it starts melting.

∴The time required before the ice starts to melt is the time required to bring the ice to 0°C.

t=\frac{Q}{q}

=\frac{19347.3}{820}

= 23.59 minutes.

Sol.: 2.

Next we need to find the time it takes before the temperature rises above freezing from the time when heating begins.

<em>Now comes the concept of Latent  heat into the play, the temperature does not starts rising for the ice as soon as it reaches at 0°C it takes significant amount of time to raise the temperature because the heat energy is being used to convert the phase of the water molecules from solid to liquid.</em>

From the above solution we have concluded that 23.59 minutes is required for the given ice to come to 0°C, now we need some extra amount of energy to convert this ice to liquid water of 0°C.

<u>We have the equation:</u> latent heat, Q_{L}= mL_{i}

Q_{L}= 0.555\times334\times10^{3}= 185370 J

<u>Now  the time required for supply of 185370 J:</u>

t=\frac{Q_{L}}{q}

t=\frac{185370}{820}

t= 226.06 minutes

∴ The time it takes before the temperature rises above freezing from the time when heating begins= 226.06 + 23.59

= 249.65 minutes

8 0
3 years ago
Study the scenario.
Otrada [13]

If no other forces act on the object, according to Newton’s first law, the spacecraft will continue moving at a constant velocity, assuming that a planet or something with large mass doesn’t cross its path. Forces are not required to continue the motion of an object on a frictionless plane at a constant rate.

7 0
3 years ago
Read 2 more answers
How can a cyclist minimize friction as he or she rides? A. by increasing the weight of the bike B. by increasing the tread on hi
jek_recluse [69]
B I believe is the answer!

Hope this helps and have a great day!!!
6 0
2 years ago
Read 2 more answers
A car is traveling at 8 m/s accelerates at 3.1 m/s^2 to reach a final top speed of 56 m/s. How much time will pass before the ca
AlekseyPX

Please find attached photograph for your answer.

Hope it helps.

Do comment if you have any query.

3 0
3 years ago
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